What is the integral of (x-1)(x^2+3) ?

The integral of (x-1)(x^2+3) is (x^4)/4+(3x^2)/2-(x^3)/3-3x+C

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integral of (x-1)(x^2+3)

\int (x - 1) (x^{2} + 3) d x

\frac{x ^{4}}{4} + \frac{3 x ^{2}}{2} - \frac{x ^{3}}{3} - 3 x + C

Solution steps

\int (x - 1) (x^{2} + 3) d x

Expand

(x - 1) (x^{2} + 3) : x^{3} + 3 x - x^{2} - 3

= \int x^{3} + 3 x - x^{2} - 3 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x^{3} d x + \int 3 x d x - \int x^{2} d x - \int 3 d x

\int x^{3} d x = \frac{x ^{4}}{4}

\int 3 x d x = \frac{3 x ^{2}}{2}

\int x^{2} d x = \frac{x ^{3}}{3}

\int 3 d x = 3 x

= \frac{x ^{4}}{4} + \frac{3 x ^{2}}{2} - \frac{x ^{3}}{3} - 3 x

Add a constant to the solution

= \frac{x ^{4}}{4} + \frac{3 x ^{2}}{2} - \frac{x ^{3}}{3} - 3 x + C

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What is the integral of (x-1)(x^2+3) ?
The integral of (x-1)(x^2+3) is (x^4)/4+(3x^2)/2-(x^3)/3-3x+C